Z30=20ww-j3 n4 Ω;j3 nj20=-j3 nO a. Z₁ = 1,95 + j1,29O b. Z₁ = 1,95 - j0,29O c. None of the alternatives are correctO d. Z 1,95 -j1,29 2O e. Z₁ = 1,95 + j0,29 0340

Answered: Image /qna-images/answer/bf678701-45ae-44b8-9e2e-4de537537d1d.jpg

Z- Σ3Ω 20 www 50 4ΩΣ -3 Ω Ο a. Zτ = 1,95 + j1,29 Ω Ο b. Ζτ = 1,95 - j0,29 Ω j2 Ω F-3 Ω 3 Ω Ο c. None of the alternatives are correct Ο d. Zτ = 1,95 - j1,29 Ω O e. Zτ = 1,95 + j0,29 Ω 34Ω

Z- Σ3Ω 20 www 50 4ΩΣ -3 Ω Ο a. Zτ = 1,95 + j1,29 Ω Ο b. Ζτ = 1,95 - j0,29 Ω j2 Ω F-3 Ω 3 Ω Ο c. None of the alternatives are correct Ο d. Zτ = 1,95 - j1,29 Ω O e. Zτ = 1,95 + j0,29 Ω 34Ω

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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Question

Z
30
=
20
ww
-j3 n
4 Ω;
j3 n
j20
=-j3 n
O a. Z₁ = 1,95 + j1,29
O b. Z₁ = 1,95 - j0,29
O c. None of the alternatives are correct
O d. Z 1,95 -j1,29 2
O e. Z₁ = 1,95 + j0,29 0
340

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